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Problems
Contests
National and Regional Contests
Argentina Contests
Argentina National Olympiad
2012 Argentina National Olympiad
2012 Argentina National Olympiad
Part of
Argentina National Olympiad
Subcontests
(6)
4
1
Hide problems
sum -1/a_b_i where a_i, b_i are min and max perfect square <= or >= i resp
For each natural number
n
n
n
we denote
a
n
a_n
a
n
as the greatest perfect square less than or equal to
n
n
n
and
b
n
b_n
b
n
as the least perfect square greater than
n
n
n
. For example
a
9
=
3
2
a_9=3^2
a
9
=
3
2
,
b
9
=
4
2
b_9=4^2
b
9
=
4
2
and
a
20
=
4
2
a_{20}=4^2
a
20
=
4
2
,
b
20
=
5
2
b_{20}=5^2
b
20
=
5
2
. Calculate:
1
a
1
b
1
+
1
a
2
b
2
+
1
a
3
b
3
+
…
+
1
a
600
b
600
\frac{1}{a_1b_1}+\frac{1}{a_2b_2}+\frac{1}{a_3b_3}+\ldots +\frac{1}{a_{600}b_{600}}
a
1
b
1
1
+
a
2
b
2
1
+
a
3
b
3
1
+
…
+
a
600
b
600
1
5
1
Hide problems
finite sequence with terms from 0-121
Given a finite sequence with terms in the set
A
=
{
0
,
1
,
…
,
121
}
A=\{0,1,…,121\}
A
=
{
0
,
1
,
…
,
121
}
, it is allowed to replace each term by a number from the set
A
A
A
so that like terms are replaced by like numbers, and different terms by different numbers. (Terms may remain without replacement.) The objective is to obtain, from a given sequence, through several such changes, a new sequence with sum divisible by
121
121
121
. Show that it is possible to achieve the objective for every initial sequence.[hide=original wording]Dada una secuencia finita con términos en el conjunto A={0,1,…,121} , está permitido reemplazar cada término por un número del conjunto A de modo que términos iguales se reemplacen por números iguales, y términos distintos por números distintos. (Pueden quedar términos sin reemplazar.) El objetivo es obtener, a partir de una sucesión dada, mediante varios de tales cambios, una nueva sucesión con suma divisible por 121 . Demostrar que es posible lograr el objetivo para toda sucesión inicial.
6
1
Hide problems
person in each square of a 2012 x 2012 board
In each square of a
2012
×
2012
2012\times 2012
2012
×
2012
board there's a person. People are either honest, who always tell the truth, or liars, who always lie. At a given moment, each person makes the same statement: "In my row there are the same number of liars as in my column." Determine the minimum number of honest people that can be on the board.
2
1
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product (11x_i^2 +12y_i^2) is perfect square
Determine all natural numbers
n
n
n
for which there are
2
n
2n
2
n
distinct positive integers
x
1
,
…
,
x
n
,
y
1
,
…
,
y
n
x_1,…,x_n,y_1,…,y_n
x
1
,
…
,
x
n
,
y
1
,
…
,
y
n
such that the product
(
11
x
1
2
+
12
y
1
2
)
(
11
x
2
2
+
12
y
2
2
)
…
(
11
x
n
2
+
12
y
n
2
)
(11x^2_1+12y^2_1)(11x^2_2+12y^2_2)…(11x^2_n+12y^2_n)
(
11
x
1
2
+
12
y
1
2
)
(
11
x
2
2
+
12
y
2
2
)
…
(
11
x
n
2
+
12
y
n
2
)
is a perfect square.
1
1
Hide problems
min and max of z if x+y+z=7 and xy+yz+zx=11
Determine if there are triplets (
x
,
y
,
z
)
x,y,z)
x
,
y
,
z
)
of real numbers such that
{
x
+
y
+
z
=
7
x
y
+
y
z
+
z
x
=
11
\begin{cases} x+y+z=7 \\ xy+yz+zx=11\end{cases}
{
x
+
y
+
z
=
7
x
y
+
yz
+
z
x
=
11
If the answer is affirmative, find the minimum and maximum values of
z
z
z
in such a triplet.
3
1
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ratio of segments wanted, incircle, circumcircle, arc midpoint related
In the triangle
A
B
C
ABC
A
BC
the incircle is tangent to the sides
A
B
AB
A
B
and
A
C
AC
A
C
at
D
D
D
and
E
E
E
respectively. The line
D
E
DE
D
E
intersects the circumcircle at
P
P
P
and
Q
Q
Q
, with
P
P
P
in the small arc
A
B
AB
A
B
and
Q
Q
Q
in the small arc
A
C
AC
A
C
. If
P
P
P
is the midpoint of the arc
A
B
AB
A
B
, find the angle A and the ratio
P
Q
B
C
\frac{PQ}{BC}
BC
PQ
.